gpt pro 5.4 attempt on Math-738

Erdős Problem 269 --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): the finite-set problem remains open, so the most promising strict advance is to isolate the multiplicity obstruction identified in Round 1, turn it into a precise irrationality criterion, and then prove a rigorous irrationality theorem for a broad infinite-prime-set regime. In addition, I explicitly re-check the unverified forum proof strategy and repair only the part that survives adversarial scrutiny.

2) Round-1 FOUNDATION USED

I use the following Round 1 items without re-proving them.

• The sequence of distinct denominators can be studied by grouping equal values of \[ L_n=[a_1,\dots,a_n]. \] Round 1 already identified the main obstruction: the original series is a weighted version \[ \sum_i \frac{c_i}{d_i} \] with multiplicities $c_i$, and these multiplicities are what prevent the simple Cantor-series tail argument from resolving the finite case.
• Round 1 also proved convergence of the series, so all rearrangements into grouped denominator levels are legitimate.

3) NEW INSIGHT / TOOL (ROUND-2)

The new tool is a prime-power frontier decomposition.

Let $P$ be any set of primes, and let \[ 1<q_1<q_2<\cdots \] be the increasing sequence of all prime powers $p^r$ with $p\in P$ and $r\ge 1$. Define \[ q_0:=1, \qquad x_i:=\mathrm{lcm}(q_1,\dots,q_i)\quad (i\ge 1), \qquad x_0:=1. \] For each $i\ge 0$, let $m_i$ be the number of $P$-smooth integers in the half-open interval $[q_i,q_{i+1})$.

Then the original series becomes \[ \sum_{n\ge 1}\frac1{L_n} = \sum_{i\ge 0}\frac{m_i}{x_i}. \] The key Round-2 criterion is:

Criterion. If $P$ is infinite and \[ \frac{m_i}{q_i}\to 0, \] then the grouped series $\sum_{i\ge 0} m_i/x_i$ is irrational.

This is strictly stronger than anything obtained in Round 1, and it pinpoints a single sharp obstruction: one only needs sufficiently small multiplicity relative to the prime-power frontier.

4) ATTACK PLAN (ROUND-2)

The exact remaining gap after Round 1 was: repeated denominators can be too numerous for the simple tail $<1/d_i$ argument.

The new plan is:

• encode denominator changes by the ordered prime powers $q_i$;
• show that the tail after a new prime enters the denominator is an integer multiple of $1/x_N$ under a rationality assumption;
• show that this integer tends to $0$ whenever $m_i=o(q_i)$;
• deduce irrationality in every situation where one can actually prove $m_i=o(q_i)$.

This directly overcomes the Round 1 obstacle in all regimes where the multiplicities are genuinely sparse relative to the frontier.

5) WORK (ROUND-2)

Lemma 5.1.

If $q_{i+1}=p^r$, then \[ x_{i+1}=p\,x_i. \] Equivalently, if $\alpha(q_{i+1})$ denotes the prime base of $q_{i+1}$, then \[ x_{i+1}=\alpha(q_{i+1})x_i. \]

Proof. Since the $q_j$ are the distinct prime powers coming from $P$, if $q_{i+1}=p^r$ then $p^{r-1}$ already occurred earlier but $p^r$ did not. Thus the $p$-adic valuation of the lcm increases by exactly $1$, while all other prime-adic valuations stay unchanged. Hence $x_{i+1}=p x_i$.

Lemma 5.2.

For every $i\ge 0$, \[ \sum_{n\ge 1}\frac1{L_n}=\sum_{i\ge 0}\frac{m_i}{x_i}. \]

Proof. Fix $i\ge 0$. For every $P$-smooth integer $u$ with $q_i\le u<q_{i+1}$, all prime powers from $P$ not exceeding $u$ are precisely $q_1,\dots,q_i$, so \[ \mathrm{lcm}\{v\in\mathbf Z_{>0}: v\le u,\ v\text{ is }P\text{-smooth}\}=x_i. \] Exactly $m_i$ values of $u$ lie in $[q_i,q_{i+1})$, so grouping equal denominators gives the formula.

Lemma 5.3.

Let $p_1$ be the smallest prime in $P$. Then for every $i\ge 0$, \[ q_{i+1}\le p_1 q_i. \] Consequently, \[ q_i\le p_1^{\,i-N-1}q_{N+1}\qquad (i\ge N+1). \]

Proof. Given $q_i$, choose the unique integer $s$ with \[ p_1^{s-1}\le q_i < p_1^s. \] Then $p_1^s$ is a prime power from $P$, it is $>q_i$, and $p_1^s\le p_1 q_i$. Since $q_{i+1}$ is the next prime power after $q_i$, we get $q_{i+1}\le p_1 q_i$. Iterating gives the second claim.

Proposition 5.4 (irrationality criterion from sparse multiplicities).

Assume that $P$ is infinite and that \[ \frac{m_i}{q_i}\to 0. \] Then \[ \sum_{n\ge 1}\frac1{L_n} \] is irrational.

Proof. Write \[ S:=\sum_{n\ge 1}\frac1{L_n}=\sum_{i\ge 0}\frac{m_i}{x_i}. \] Assume for contradiction that $S=a/b\in\mathbf Q$ in lowest terms.

Because $P$ is infinite, infinitely many of the $q_j$ are actual primes. Choose $N$ so that $q_{N+1}$ is a prime. Define the tail \[ T_N:=\sum_{i\ge N+1}\frac{m_i}{x_i}. \] Since $x_i\mid x_N$ for $i\le N$, we have \[ I_N:=b x_N T_N = ax_N-bx_N\sum_{i=0}^{N}\frac{m_i}{x_i} \in \mathbf Z_{>0}. \] So $I_N$ is a positive integer.

Now let \[ M_N:=\sup_{i\ge N+1}\frac{m_i}{q_i}. \] Since $m_i/q_i\to 0$, we have $M_N\to 0$ as $N\to\infty$.

Using Lemma 5.1, \[ \frac{x_N}{x_i} = \frac1{\alpha(q_{N+1})\alpha(q_{N+2})\cdots \alpha(q_i)}. \] Hence \[ I_N = b\sum_{i\ge N+1} \frac{m_i}{\alpha(q_{N+1})\alpha(q_{N+2})\cdots \alpha(q_i)} \le b M_N \sum_{i\ge N+1} \frac{q_i}{\alpha(q_{N+1})\alpha(q_{N+2})\cdots \alpha(q_i)}. \] Because $q_{N+1}$ is a prime, $\alpha(q_{N+1})=q_{N+1}$. By Lemma 5.3, \[ \frac{q_i}{\alpha(q_{N+1})\cdots \alpha(q_i)} \le \frac{p_1^{\,i-N-1}}{\alpha(q_{N+2})\cdots \alpha(q_i)}. \] Let $p_2$ be the second-smallest prime in $P$, and let \[ n_i:=\#\{j\in\{N+2,\dots,i\}: \alpha(q_j)>p_1\}. \] Since each factor $\alpha(q_j)$ is either $p_1$ or at least $p_2$, \[ \alpha(q_{N+2})\cdots \alpha(q_i) \ge p_1^{\,i-N-1-n_i}p_2^{n_i}, \] so \[ \frac{q_i}{\alpha(q_{N+1})\cdots \alpha(q_i)} \le \left(\frac{p_1}{p_2}\right)^{n_i}. \] Therefore \[ I_N\le b M_N \sum_{i\ge N+1}\left(\frac{p_1}{p_2}\right)^{n_i}. \] It remains to bound the last series uniformly in $N$.

Set \[ \ell:=\left\lceil \frac{\log p_2}{\log p_1}\right\rceil. \] For every $x>0$, the next power of $p_2$ after $x$ is at most $p_2 x\le p_1^{\ell}x$. Thus every interval $(x,p_1^{\ell}x]$ contains a power of $p_2$. In particular, it is impossible to have $\ell+1$ consecutive terms of the prime-power sequence $(q_i)$ all with base prime $p_1$: the first and last of those terms would differ by a factor at most $p_1^{\ell}$, but there would have to be a power of $p_2$ strictly between them. Hence $n_i$ can stay constant for at most $\ell+1$ consecutive indices. Therefore, for each $k\ge 0$, \[ \#\{i\ge N+1: n_i=k\}\le \ell+1. \] So \[ \sum_{i\ge N+1}\left(\frac{p_1}{p_2}\right)^{n_i} \le (\ell+1)\sum_{k\ge 0}\left(\frac{p_1}{p_2}\right)^k =:C(P)<\infty, \] with $C(P)$ independent of $N$.

Hence \[ 0<I_N\le b C(P) M_N\to 0. \] But $I_N$ is a positive integer, contradiction. Therefore $S$ is irrational.

Corollary 5.5.

Assume $P$ is infinite and one of the following holds.

• (i) $q_{i+1}-q_i=o(q_i)$ for the ordered prime-power frontier;
• (ii) $\sum_{p\notin P}1/p=\infty$.

Then \[ \sum_{n\ge 1}\frac1{L_n} \] is irrational.

Proof. In case (i), \[ m_i\le q_{i+1}-q_i=o(q_i), \] so Proposition 5.4 applies.

In case (ii), let $h_i:=q_{i+1}-q_i$. Every integer counted by $m_i$ lies in an interval of length $h_i$ and has no prime factor outside $P$. Brun's upper-bound sieve gives \[ m_i\ll h_i\prod_{\substack{p\le h_i\\ p\notin P}}\left(1-\frac1p\right). \] If $h_i=o(q_i)$ then we are done. Otherwise there is some $\varepsilon>0$ along an infinite subsequence with $h_i\ge \varepsilon q_i$, so $h_i\to\infty$ there; because $\sum_{p\notin P}1/p=\infty$, the Euler product above tends to $0$, and thus along that subsequence \[ m_i=o(h_i)=o(q_i). \] Therefore $m_i=o(q_i)$ in all cases, and Proposition 5.4 applies.

Round-2 consequence.

The proposition and corollary give a complete irrationality proof for a broad infinite-$P$ regime, including for example the case where $P$ is the set of all primes.

6) ADVERSARIAL VERIFICATION

There is one important correction to make relative to the unverified forum proof sketch.

• The step "$\sum_{p\notin P}1/p<\infty$ implies every interval $(x,(1+\varepsilon)x]$ eventually contains a prime of $P$" is not justified by the argument sketched on the forum page. Convergence of the complementary reciprocal sum alone does not obviously force this multiplicative-syndetic property.
• The proof written above therefore does not claim irrationality for every infinite $P$. What it proves is the precise conditional theorem in Proposition 5.4, together with the rigorous Corollary 5.5 covering two broad classes of infinite sets.
• Every use of positivity and integrality in Proposition 5.4 has been checked: $I_N>0$ because the tail is positive, and $I_N\in\mathbf Z$ because $x_i\mid x_N$ for all $i\le N$.
• The boundedness of the auxiliary series was also checked carefully: the only input is that powers of $p_2$ occur regularly on the multiplicative scale, and Lemma 5.3 converts this into a uniform bound on the number of indices with a fixed value of $n_i$.

So the new argument is rigorous as stated, but it still leaves open the genuinely difficult remaining infinite-$P$ cases, and of course the original finite-$P$ problem.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

I proved a new irrationality criterion for the grouped series and obtained a rigorous irrationality theorem for all infinite prime sets $P$ satisfying $m_i=o(q_i)$, in particular for every infinite $P$ with either $q_{i+1}-q_i=o(q_i)$ or $\sum_{p\notin P}1/p=\infty$. I also identified a concrete gap in the previously cited unverified "all infinite $P$" argument. The original finite-$P$ problem remains open.

8) COMPLETION ESTIMATE

COMPLETION: 45%

9) REFERENCES

• T. F. Bloom, Erdős Problem #269, checked 2026-03-07.
• T. F. Bloom, Discussion thread for Erdős Problem #269; in particular the comment of Steve Fan (11 Oct 2025), checked 2026-03-07. Used as a heuristic starting point only; the Round-2 proof above is the corrected version actually verified here.